The triangle is equilateral. The points are respectively on the sides. We know that there are exactly isosceles triangle from the triangles. How many possible values are there for ?
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To prove existence of a the divisions with a given number of isosceles triangles, all we have to do is show an example with that number, as is done above for numbers 4, 2, 1, and 0.
What is left is to prove non-existence of a division with exactly 3 isosceles triangles. At least two of these would have to be triangles with one of their vertices being A, B, or C. That means they would have to have at least one 6 0 ∘ angle, but an isosceles triangle with one angle 6 0 ∘ is an equilateral triangle. They could not be the same size, otherwise all the triangles would become the same and there would be 4 triangles, not 3.
This means that the third one could not be also in a corner, as that one has to have two different sides next to a 6 0 ∘ angle, so will not be isosceles. But the triangle XYZ will also have a 6 0 ∘ angle, because one of its vertices will be between the two equilateral triangles, and also would have to have two different size sides making up that angle.
So there are only 4 possibilities - 0, 1, 2, and 4.